Question

A particle moves along a straight line and at a distance x from a fixed origin O on the line. Its velocity is proportional to $\sqrt {\dfrac{{\beta - \alpha }}{x}} .$ Then, its acceleration is
A. directed towards O and is proportional to $1/x.$
B. directed towards O and is proportional to $1/{x^2}.$
C. directed towards O and is proportional to x.
D. directed away from O and is proportional to ${x^2}.$

Answer 1

Hint:As velocity is proportional to $\sqrt {\dfrac{{\beta - \alpha }}{x}} $ and we have to find acceleration, so basic concept of velocity and acceleration and inter-relation between then is used.
Formula used:
A. Acceleration, $a = \dfrac{{vdv}}{{dr}}$
Where v is the velocity
du is a small change in velocity with small change in distance dx.

Complete step by step answer:
Now, we know that velocity is time rate of change of displacement i.e.
$v = \dfrac{{dx}}{{dt}}.......\left( 1 \right)$
where v is the velocity
Now, acceleration is the time rate of change of velocity. It can also be written as
Acceleration, $a = \dfrac{{dv}}{{dr}}.......\left( 2 \right)$
In order to find relation between acceleration with velocity as a function of distance x,
Multiply and divide equation (2) by dx, we get, acceleration, $a = \dfrac{{dv}}{{dt}} \times \dfrac{{dx}}{{dt}}$
$a = \dfrac{{dv}}{{dx}} \times \dfrac{{dx}}{{dt}}$
$a = \dfrac{{dv}}{{dt}} \times v$ (from equation 1)
Now, as in the question, velocity is given to be proportional to $\sqrt {\dfrac{{\beta - \alpha }}{x}} $ i.e.
$v\,\,\alpha \,\sqrt {\dfrac{{\beta - \alpha }}{x}} \; \Rightarrow v = k\sqrt {\dfrac{{\beta - \alpha }}{x}} .......\left( 4 \right)$
Where to is proportionality constant. Also $\beta $ and $\alpha $ are treated as constant. Differentiating v with respect to x, we get
$\dfrac{{dv}}{{dx}} = \dfrac{d}{{dx}}\left( {k\sqrt {\dfrac{{\beta - \alpha }}{x}} } \right) = k\sqrt {\beta - \alpha } \,\,x\,\,\dfrac{{ - 1}}{2}\,\,\dfrac{1}{{x\sqrt x }}$
$\dfrac{{dv}}{{dx}} = \dfrac{{ - k\sqrt {\beta - \alpha } }}{{2x\,\,\sqrt x }}........\left( 5 \right)$
Put equation (4) and (5) in equation (3),
We get
$a = \dfrac{{ - k\sqrt {\beta - \alpha } }}{{2x\,\,\sqrt x }} \times k\sqrt {\dfrac{{\beta - \alpha }}{x}} $
$ \Rightarrow a = \dfrac{{ - {k^2}\left( {\beta - \alpha } \right)}}{{2x \times x}}$
$ \Rightarrow a = \dfrac{{ - {k^2}\left( {\beta - \alpha } \right)}}{{2{x^2}}}$
As $\beta ,\alpha ,$k,z all are constant. So, acceleration, $a\alpha \dfrac{{ - 1}}{{{x^2}}}$
So, acceleration is directed opposite to position that is it is directed towards origin and us proportional to $1/{x^2}.$

Hence, the correct option is B.

Note: While solving $\dfrac{{dv}}{{dx}},$ basic mathematical derivation is used. Which is basically
$\dfrac{d}{{dx}}\left( {\dfrac{1}{{\sqrt x }}} \right) = \dfrac{d}{{dx}}\left( {{x^{ - 1/2}}} \right)$
$ = \dfrac{{ - 1}}{2}{x^{ - 1/2 - 1}}$
$ \Rightarrow \dfrac{{ - 1}}{2}{x^{ - 3/2}}$
$ \Rightarrow \dfrac{{ - 1}}{{2x\sqrt x }}$
Also, as x is distance from origin and a is proportional to negative of $1/{x^2}$ so directed towards origin.